The following is an exercise from D. Robinson: A Course in the Theory of Groups.
Let $G$ be a $k$-transitive permutation group of degree $n$ which is neither alternating nor symmetric. Assume $k > 5$. Prove that $(n-k)! \ge 2n$. Deduce that $k \le n-4$.
If $G$ is $k$-transitive, then $n(n-1)\cdots (n-k+1)$ divides $|G|$. Now looking at the first inequality it seems to be related to the order $\frac{(n-k)!}{2} = |A_{n-k}|$ of the alternating group on $n-k$ symbols. Denote by $(G_{\alpha})_G = \bigcap_{g\in G} G_{\alpha}^g$ the normal core of $G_{\alpha}$, as $|G : (G_{\alpha})_G| \ge |G : G_{\alpha}| = n$, if I can embed somehow the factor group $G / (G_{\alpha})_G$ into $A_{n-k}$ the result would follow. This are just some thoughts and I have no idea how to proceed, on what set of size $n-k$ should $(G/G_{\alpha})_G$ act such that it only produces even permutations I do not know. I have no other idea, so I am stuck. Does anyone know how to solve it?